Probability Experiments, Outcomes, Samples, Spaces, Events

Hi, welcome to www.eudcator.com, my name is Will Murray and0000

we are going to be starting the probability lectures today.0004

Thank you very much for joining us, we are going to jump right on in.0007

We are going to learn some terminology in this first lesson.0010

We are going to learn about experiments and outcomes and sample spaces and events.0012

Let us get right started with that.0018

The words that we are throwing at you are all used to describe probability experiments.0020

Let me first got introduce the word experiment itself.0027

An experiment is a process leading to exactly one of various possible outcomes.0030

We will see lots of examples of this as we go through the lectures.0037

But some example to keep in mind, a very simple example would be flipping a coin, or you roll a dice,0040

and you draw a card from a deck.0048

Those are the kinds of things that we can run in an experiment in exactly one of the possible outcomes will occur.0050

Having said that, I should say what an outcome is one of the things that can happen in an experiment.0057

Outcomes are also known as simple events or sample points.0066

All of those words are synonymous, we use them interchangeably.0070

A simple event, a sample point, or outcome.0073

If you put all the possible outcomes together that is called the sample space.0077

The sample space is essentially the set of all things that can happen in an experiment.0083

If you are going to flip one coin, for example the sample space is head and tail0087

because those are the 2 things that can happen.0093

You can get a head or a tail.0094

The sample space is also known as probability space.0097

Those words were used interchangeably, probability space or sample space.0101

If you are a studying probability, with your own textbook they might use the word sample space,0106

they might use the word probability space, those mean the same thing.0111

Finally, an event is a subset of the sample space that is a set of some of the outcomes.0115

An event means for example, if we are going to pick a card from a deck of cards,0122

the event could be that we get a spade because there are 52 cards in a standard deck and 13 of them are spades.0128

We collect all those 30 outcomes together and call them an event.0137

We often use the variable S to describe the set of all possible outcomes so the sample space.0142

We often use the variable A or sometimes A and B when we have multiple ones to describe different events in that sample space.0149

We will talk about the probability of an event A occurring when we run an experiment.0158

Let us figure out how we compute the probabilities.0164

When we do that, if we have an event is assuming that these are things that we can count and everything is nice and finite,0168

first of all we count the number of outcomes in the sample space S.0179

We count all the possible things that can happen in an experiment.0184

If you are flipping a coin, the only 2 things that can happen are head and tail.0188

You just count the head and the tail, you get 2 possible outcomes.0193

If you are drawing a card from a deck, there are 52 different things that can happen0196

because you can get the ace of spades or the 2 spades, up to the king of spades and so on.0200

There are 52 different cards that you might draw.0205

Those are all the outcomes.0208

To calculate the probability of an event, you want to count the number of outcomes in that event.0210

For example, if you are drawing a card from a 52 card deck and0217

you want to calculate the probability of getting a spade, P of getting a spade when you draw a single card from a deck.0223

There are 52 possible outcomes, there are 52 possible cards that you can draw, and 13 of them qualify as being in that event.0237

There are 13 possible outcomes which will give you spades.0248

Of course, you can reduce the fraction 13/52 and you find that the probability of getting a spade is ¼.0253

That is not very surprising.0261

Because of this of nature a probability, you are counting things all over the place.0264

There is a lot about counting outcomes, counting events.0269

A great deal of early focus in probability course is it is going to have can have a common notarial flavor,0274

when we try to count different events and count different sample spaces.0281

You will see that some of the problems that we study today and in some of the other early lectures0285

are all about counting different things, different complicated counting techniques.0290

You get the hang of it as we get into it, as we start practice it a little more.0296

Our first example is to roll 2 dice and to see if the sum showing on the 2 dice is 10.0302

Remember, these are standard dice so they are labeled 1 through 6 on the 6 sides.0308

I want to see if the sum is going to be a 10 between the 2 dice.0313

We want to identify the experiment.0318

We want to list all the possible outcomes and the event that we are interested in,0321

which means the subset of outcomes that would give us 10.0325

We want to find the probability of the event.0330

The probability that we are going to get a 10 when we roll 2 dice.0331

This is a pretty standard probability problem.0336

The experiment here is just roll the 2 dice.0339

We are rolling the dice, roll the 2 dice.0347

The outcomes are all the different possible ways that the 2 dice can call up.0355

And it is very helpful here if you have a mental image of the dice being distinct.0365

Maybe think of one of them as being red and one of them as being blue.0372

Think of you are rolling a red dice and a blue dice.0376

Let us think of all the different combinations that can arise there.0379

The red dye could be 1, 2, 3, 4, 5, and 6.0384

The blue dice could be 1, 2, 3, 4, 5, and 6.0391

They are all in the different possible combinations of the 2 dice there.0401

The red could be 1, the blue could be 1.0405

The red could be 1 and the blue could be 2.0409

The red could be 1 and the blue could be 3, and so on.0414

There are 6 possible combinations in that first roll there.0419

There are going to be 6 possible combinations in the 2nd roll, 3rd roll, 4th roll, 5th roll, all the way up to0423

you could get 6 on both sides.0429

You can get red dice being 6 and the blue dice being 6.0431

There are 36 possible outcomes in the set there.0436

Our sample space here is all those possible outcomes.0443

We can have a red 1 and a blue 1.0447

We can have a red 1 and a blue 2.0451

Let me see if I can make it look more blue there.0461

All the way up to a red 6 and a blue 6.0465

We have 36 possible outcomes in this experiment, 36 possible combinations of numbers on the 2 dice.0474

That is our sample space, all the possible outcomes.0485

We want to identify the event we are interested in.0488

The event here is that the 2 dice sum up to 10.0492

Let us think about which of those possible outcomes will give us a sum of 10.0497

You can get a red 4 and a blue 6.0503

You can get a red 5 and a blue 5, because those that up to 10.0508

Or you can get a red 6 and a blue 4, because those also add up to 10.0515

3 of those combinations between red and blue add up to 10.0522

That is the event we are interested in.0529

We now know what our event is and we now know what our sample space is.0533

Let us figure out what the probability is.0538

The probability of event A is the number of outcomes in A ÷ the total number of outcomes in our sample space.0541

The total number of things that can happen, outcomes in S.0560

And we calculated both of those, we looked at event A and we saw that there were 3 outcomes that will give you a sum of 10.0568

In S, there are 36 possible outcomes and that simplifies down into 1/12.0576

That is the probability of rolling a 10, when you roll a 2 dice together.0585

That is our final probability there.0594

Just to recap here, what we did was we identified our experiment which was rolling the 2 dice.0600

We listed all the possible outcomes, all the possible outcomes was this chart right here.0607

There are 36 possible outcomes, possible combinations of what might be showing on the red dice,0613

what might be showing on the blue dice.0619

I did not actually list all 36 because that will be too cumbersome but we know there will be 36 of them.0622

And then we have to identify the event which is that 2 dice add up to 10.0628

There are 3 possible outcomes, 3 of those combinations add up to 10.0636

We mention right here a place that often confuses students when I’m first starting to study probability,0640

people wonder that it looks like 4 or 6 got listed twice and 5 -5 only got listed once.0646

That is true, that is because there are 2 different ways you can get a 4 – 6.0654

You get a red 4 and blue 6 or you can get a red 6 and blue 4.0659

That will get listed twice.0664

5 and 5, there is only one way you can get it which is to get 5 on both of the dice.0666

There really are 2 ways to get a 4 and 6.0673

There is only one way to get a 5 and 5.0676

In fact, this is even reflected when people roll dice in casinos in the game craps,0679

people talk about getting 10 the hard way which means getting a 5 and 50685

because that is harder than getting a 4 and 6 because there are 2 ways to get a 4 and 6.0690

We listed all of them, we found 3 outcomes there.0697

The probability of our event is just the number of outcomes in the event and the number of outcomes ÷0700

the number of outcomes in our sample space.0707

There are 3 outcomes in our events, we counted those.0710

There are 36 in our sample space, 3/36 reduces down to 1/12.0714

Let us keep going.0721

In our second example, we are going to flip a coin 3 times and see if we get exactly 2 heads.0723

And again, we want to identify the experiment, all the possible outcomes in the event we are interested in.0729

We want to find the probability of the event.0735

The experiment here is flipping the coin.0739

The experiment is to flip the coin 3 times, all the possible outcomes0742

or all the possible combinations of heads and tails that we can get.0755

The possible outcomes here, I’m just going to try to list them systematically.0761

Let me call this S because this is the sample space that we are listing right now.0768

S is we can get head, head, head.0772

We can get a head, head, and a tail.0778

We can get a head, a tail, and then a head.0781

A head, then a tail, then another tail.0785

We can get a tail-head-head.0789

We can get tail-head-tail.0793

Tail-tail-head or a tail-tail-tail.0796

There are 8 possible outcomes here and that is really coming from the fact that 2 × 2 × 2 is 8.0802

Because each time you flip the coin, there are 2 different things that can happen.0812

Then you multiply those possibilities as you go along, so you get 8 possible combinations of flips there.0816

Those are the outcomes that we are interested in, that is the entire sample space.0823

The event that we are interested in is that we get exactly 2 heads.0830

I want to figure out which of those outcomes satisfies that criterion.0844

Exactly 2 heads would mean certainly HHH, it does not but HHT does and HTH does, head- tail- head.0849

Down there I see a tail-head-head0860

Those are the only ones there are that have exactly 2 heads.0866

Our event just has those 3 possible outcomes that satisfy the requirement.0872

And then finally, we want to find the probability of the event.0878

The probability of A is the number of outcomes in A ÷ the total number of outcomes0881

in our sample space, in our probability space, outcomes in S.0894

In A, we had 3 possible outcomes and in S we had 8.0904

That does not reduce so our final answer there is 3/8.0910

That is the probability of getting exactly 2 heads, when we flip a coin 3 times.0914

Let we make sure that all the steps there are clear, we want to identify the experiments.0920

Identifying the experiment is just flipping the coin 3 times.0925

We wanted to list all the possible outcomes.0929

I listed them in order here, I kept track of the first flip, the second flip, and the third flip.0933

There are 8 possible outcomes.0938

That came from 2³ because there are two things that can happen on the first flip × 2 things0943

that can happen on the second flip × 2 things that can happen on the third flip.0951

Exactly 2 heads, I just scan through that list of outcomes and I have identified the ones that have exactly 2 heads.0955

The first one has 3 heads, that does not count.0962

The second one and the third one both have exactly 2 heads.0964

The 5th one here has exactly 2 heads.0969

And that is why I took those and put them into now the set, describing the event that we are interested in.0972

The probability of that event is just the number of outcomes in A which is 3 ÷ the number of outcomes in S which is 8.0980

That is where we got the probability being 3/8 there.0989

For our next example, we are going to be drawing cards from a deck.0995

We are going to take a standard 52 card deck and we are going to keep drawing cards until we find the ace of spades.1000

The question is what is the probability that it will take us between 20 and 30 cards,1011

including 20 and 30, that is what inclusive means.1015

When it says inclusive, it means you include those starting and ending values.1018

It says identify the experiment, all the outcomes, and the event that we are interested in.1024

The experiment here is drawing the cards.1031

We are drawing the cards, do not know how long it is going to last because we might get the ace of spades on the very first pick.1037

Or if we get unlucky, we might have to draw 51 cards and then get the ace of spades on the very last pick.1047

The set of possible outcomes really depends on where we get the ace of spades.1054

We could get the ace of spades on the first pick, if we are very lucky.1061

We could get it on the second pick, that would still be pretty lucky if we get the ace of spades on the second pick,1070

anywhere up to the 52nd pick.1076

It would be our worst case scenario, if we got the ace of spades as the very last card that we draw.1080

Remember, we said without replacement which means after we draw a card, we are not putting it back in the deck.1087

We know for sure that eventually I’m going to find the ace of spades.1094

If we are putting the cards back in the deck, I might potentially never find the ace of spades1098

because I could keep drawing forever and just keep drawing new cards.1102

The event here is that we are going to draw between 20 and 30 cards.1106

Our event would be that we get the ace of spades on the 20th pick or possibly on the 21st pick,1112

or all the way up to the 30th pick.1130

That is our event there, that is our A.1135

Generally, use A and B for events.1140

We use the letter S for the sample space.1143

The probability of A, that is what we are really trying to calculate here,1146

is the number of outcomes in A ÷ the number of outcomes in S.1150

We just need to calculate that.1169

The number of outcomes in S, we already said there is 1 pick up to 2 picks up to the 52nd pick.1171

There are 52 possible outcomes in S.1177

If we look at A, 20, 21, up to 30, there are 11 of those outcomes because we are counting the 20 and 30 on both ends.1181

It is 20, 21, 22,23, 24, 25, 26, 27, 28, 29, 30.1190

It is 11 outcomes in the event A.1195

We actually have 11 outcomes not 10.1200

It seems like there are 10 but there is extra 1 on the end.1203

That is our probability of drawing the ace of spades somewhere between the 20th card and the 30th card.1207

Let me go back over that and make sure everything is clear.1218

The experiment here is drawing the cards.1221

We are drawing the cards one by one.1223

We might just draw a single card.1225

We might end up drawing all the way to the 52nd card because we stop whenever we get the ace of spades.1226

And we are not putting cards back in the deck, that is the very key point to mention here.1233

The outcomes are that we could get the ace of spades on the first pick, the second pick, all the way up the 52nd pick.1239

That is 52 possible outcomes in our sample space.1245

The event that we are interested in, according to the problem is that we might get the ace of spades1250

on the 20th pick, the 21st pick, or all the way up through the 30th pick.1257

There are 11 of those possibilities that we are interested in to satisfy the requirements of the problem.1263

When we are calculating the probability, we divide those numbers together.1269

The number of outcomes in the event we are interested in ÷ the total number of outcomes just gives us 11/52.1273

In example 4, it is the same experiment as before.1283

We are going to roll 2 dice, I think we saw that in example 1.1288

This time want to see if the sum is a prime number.1292

Again, we are going to identify the experiment, all the outcomes in the event we are interested in and1295

we are also going to find the probability of the event.1300

The experiment, just like before, is to roll 2 dice.1303

I think this was in example 1 before.1309

Yes, that was example 1.1312

Our experiment is the same thing, we are going to roll 2 dice but now the outcomes, those are also the same as before.1314

Those are all the possible combinations you can get when you roll 2 dice.1329

And again, it is very helpful to think of the dice as being red and blue.1333

You can get a 1, 2, 3, 4, 5, and 6 on the blue dice.1337

You can also at the same time get 1, 2, 3, 4, 5, and 6, on the red dice.1345

In terms of combinations, you can have red 1 blue 1.1352

You can have red 1 blue 2, and so on.1356

There are 36 possible combinations all the way down to red 6 blue 6.1361

That is my set of outcomes, that is my sample space there, 36 possible combinations there.1375

This is S and we want to identify the event here.1382

The event here is the sum showing on the 2 dice is a prime number.1386

Let us figure out which of those combinations could give us a prime number.1393

I’m going to use a new screen for this because it takes a little bit of calculation.1397

The event here is, I’m going to call it A.1404

The set of combinations for which the sum is a prime number.1410

Let us remember our prime numbers.1421

The prime numbers between 2 and 12, because those are the only possible combinations you can get by rolling 2 dice.1424

You can get a 2, 2 is prime, 3 is prime, 5 is prime, 7 is prime, and 11 is prime.1431

Those are all the possible totals we are looking for.1440

I want to figure out which combinations will give me those possible totals.1444

To get a 2, I could roll a 1 on the red dice and 1 on the blue dice.1450

That will give me a total of 2.1456

That is the only possible combination that does give me a total of 2.1457

To get a 3, I can get 1 on the red dice, 2 on the blue dice.1461

Or I can get a 2 on the red dice and a 1 on the blue dice.1468

To get a 5, I could roll a red 1 blue 4.1474

I could roll a red 2 blue 3.1482

That was supposed to be blue.1487

I could roll a red 3 blue 2.1492

Or I could roll a red 4 blue 1.1498

There are 4 possible combinations that would add up to 5.1505

For 7, I could roll a red 1 blue 6.1509

I think I’m going to stop switching colors back and forth because it is slowing me down here.1516

I can roll a red 1 blue 6, red 2 blue 5, red 3 blue 4, red 4 blue 3, red 5 blue 2, or red 6 blue 1.1522

There are 1, 2, 3, 4, 5, 6 combinations that can give me a total of 7.1535

To get 11, I could roll a red 5 and a blue 6, or a red 6 and a blue 5.1539

Those are the ways that I can get 11.1552

There is a very nice way to keep track of these.1553

Instead of trying to list all of these and not being really sure if you got them all,1556

let me remind you of what the sample space look quite.1560

For blue, we can get 1, 2, 3, 4, 5, 6.1564

For red, we can get 1, 2, 3, 4, 5, and 6.1570

For red, we can get a red 1 blue 1.1578

We can get a red 2, blue 1.1582

We can get a red 1 blue 2, and so on.1587

We can fill in the rest of this chart but when we are trying to list totals,1592

if we look at the total of 2, let me circle that in green here.1598

The only way to do it is by getting a 1-1, that is a total of 2.1604

If we want a total of 3, we can get 2-1 or a 1-2.1608

Those are the ways that you get a total of 3.1613

A total of 5 would be, you ca get a red 1 blue 4.1617

I’m sorry, I wrote that in the wrong place there.1628

That would be a red 4 blue 1.1631

You can get a red 3 blue 2.1636

You can get red 2 blue 3, or a red 1 blue 4.1641

What you get here to get a total of 5 is the diagonal stripe on the table.1650

It is 2 stripes below the stripe for 3 because 5 is 2 bigger than 3.1660

To get a 7, you will get a diagonal stripe on the table.1666

2 places below, you are going to get 6 outcomes across this diagonal stripe.1670

There will be 6 outcomes in there, dividing the blue can add up to 6.1681

If I need to get 11, you can get a red 6 blue 5 or a red 5 blue 6.1687

There is a shorter diagonal stripe there, this is to get 11 and this one is to get 7.1699

We can look at these diagonal stripes and the lengths of the stripes to calculate the total probability of this event.1708

We just add up all the lengths of those stripes.1715

That is kind of a good way to check that we have not overlooked any possibilities here.1718

That is our event, let us calculate the probability now.1724

The probability of A is equal to the number of outcomes in A ÷ the number of outcomes in S.1727

I need to count out the total number of outcomes in A.1748

If I look at 2, I see 1 possible outcome.1751

If I look at 3, I see 2 possible outcomes.1754

5, I see 4 possible outcomes.1759

7, I see 6 possible outcomes.1762

11, I see 2 possible outcomes.1764

Again, I can check these numbers by looking at the lengths of the stripes over here.1767

7 has a stripe of 6.1773

5 has a stripe of 4.1775

3 has a stripe of 2.1777

2 has a stripe of 1.1779

11 has a stripe of 2.1780

Those correspond to these numbers here.1783

And if I just add all of those numbers up, I see 2 + 6 is 8 + 4 is 12 + 2 is 14 + 1 is 15.1787

I get 15.1801

And then remember, the total number of outcomes in S, that is the total size of this chart which was 36.1803

It looks like 15 and 36, I can reduce that if I take a 3 out of each one.1809

It reduces down to 5/12, that is my probability.1813

If I roll 2 dice, what is the probability of getting a prime number?1818

The answer is 5 out of 12.1822

Let me remind you of everything we did there.1826

First, I made a chart of all the possible outcomes.1830

I did not fill them all in because it would have taken too long and I got the idea of what the shape would be.1833

There are 36 possible outcomes depending on the 6 possibilities for the red dice and the 6 possibilities for the blue dice.1841

I identified my event which is that the sum was prime, which means the sum has to be 2, 3, 5, 7.1850

I have looked at my chart and I try to list all the outcomes that would lead to the sum being 2, 3, 5, or 7.1857

That is where I got 1 -1 for 2.1864

2 possibilities adding up to 3, 4 possibilities adding up to 5, 6 possibilities adding up to 71867

and then just 2 possibilities adding up to 11.1874

To find the probability, I need to count up all those possibilities.1877

I have added up all those numbers, that is what I was doing over here.1881

Add them all up to get to 15, that is where this 15 comes from,1884

and then divide by the total number of outcomes in the sample space which is 36.1889

15/36 reduces down to 5/12.1894

That is how we got that probability.1898

For example 5, it is a little bit different from some of the others.1903

We are going to flip a coin repeatedly until we get a head and1906

I want to calculate the probability that we are going to flip at least 3 times.1910

The idea is we keep flipping this coin over and over again.1915

As soon as we see a head, we stop, and then that is the number of flips that we done in total.1918

What is the probability that that number will be 3 or greater?1923

Let us identify the experiment and all the outcomes and then find the probability of that event.1930

The experiment here is just flipping a coin over and over again.1937

And then as soon as we see the first head, we stop.1951

The outcomes, this is a little different from all of our example so far1958

because what can happen here is we could get a head right away on the first flip.1965

Our first possible outcome is we get a head right away or we might get a tail on our first flip1973

and then get a head and then would stop.1980

Or we might get a tail and then another tail and then get a head and then we would stop.1982

We might get 3 tails and then get a head and then we would stop.1988

And we do not really know how long this is going to go.1994

There is no assuming on how long this is going to go.1997

It could go on forever.2000

I cannot just list all these outcomes because there are infinitely many.2002

It could go on indefinitely.2006

The event that we are looking for is that we have at least 3 flips.2008

That is A is the set, a single flip getting us a head will be qualified.2018

A tail and a head will not qualify.2026

It would be a tail- tail-head, that would be 3 flips.2027

Tail-tail-tail-head, that would be more than 3 flips.2031

A tail-tail-tail-tail-head, that would be more than 3 flips.2036

It is anything that has 3 flips or more.2041

We want to identify the probability of that event.2044

Unfortunately, we cannot do this one by counting.2049

Because we have an infinite number of outcomes in our event,2053

we have an infinite number of outcomes that can happen from the experiment.2056

We can no longer use our counting formula from before.2061

Instead, let me introduce a new way of calculating it.2065

Our way of calculating the probability is we can add up the probabilities of each of these individual outcomes in our event.2072

It is the probability of tail-tail-head + the probability of tail-tail-tail-head + the probability of 4 tails2084

and then a head, and so on.2096

This is going to be an infinite sum.2100

We are going to have to add all these up forever.2102

We have to do something clever at some point.2106

Tail-tail-head, to get a tail-tail-head, there is a ½ chance that you get a tail right away and2108

then there is ½ chance that you get another tail and then there are ½ chance that you get a head.2117

Put those together and you get 1/8.2123

A tail-tail-tail-head, there are 50% chance of getting a tail, 50% chance of getting another tail,2126

50% chance of getting another tail, 50% chance of getting a head after that.2132

So it is ½ × ½ × ½ × ½, I you put all of those together you get 1/16.2136

Here you will be multiplying together ½⁵ which is ½⁵ is 32.2146

We have to add all those up, 1/8 + 1/16 + 1/32.2158

This is an infinite series.2168

Now, you have to remember something from your calculus 2.2170

If you do not remember how to deal with infinite series,2174

we got some great lectures here on www.educator.com on the second semester calculus, calculus level 2, taught by a great instructor.2177

And there is a lot of work in there in adding up infinite series.2187

Here we are going to apply that.2191

This is actually a geometric series.2192

And with a geometric series, you need to identify the common ratio.2198

Here the common ratio, what we are doing from each term to the next is we are multiplying by ½.2202

Each term is half as big as the previous term.2209

The important thing here is that ½ is less than 1 because that tells us when a geometric series converges,2212

it converges when the common ratio is less than 1.2220

It does converge and we learn back in those lectures on calculus 2, the sum of an infinite geometric series,2224

I’m going to write it in words, it is the first term or 1 - the common ratio.2232

In this case, our first term is that 1/8 right there.2245

It is 1/8/ 1 - the common ratio which is ½.2249

1/8/ 1 -1/2 which is 1/8/ ¼.2255

Do a flip here and we get 4/1 × 1/8.2261

I’m sorry, 1 - ½ is not ¼, it is ½.2268

When we do the flip there, it is 2/1 × 1/8 which is ¼.2273

That is our probability of having to flip this coin at least 3 times in order to see the first head.2284

I want to go back and calculate this in different way.2294

We will see a different way to solve this problem.2296

I want to do that on the next page.2298

Let me recap how we calculated it using this method first.2299

Our experiment was just flipping a coin here.2305

The possible outcomes are all the possible strings of heads and tails that we can see.2309

Remember, we stop as soon as we see the first head.2316

If we get a head right away, we stop.2318

If we get a tail, we keep flipping and then we get a head, we stop.2320

If we get 2 tails and then a head, we stop after that head.2325

If we get 3 tails and then head, we stop after that head, and so on.2328

Those are all the possible outcomes.2332

The event we are interested in was that we flip at least 3 times.2334

I listed all the outcomes that have 3 or more flips.2339

To find the probability of that, we cannot count these things anymore because there is infinitely many of them.2345

Instead, we have to find the probability of each one.2351

The probability of tail-tail-head is 1/8 because it is ½ × ½ × ½.2355

Tail-tail-head is 1/16 because you have to predict the flips 4 times in a row to get that right.2362

Here, tail-tail-tail-head, you have to predict 5 flips in a row.2370

You have 1/2 chance of each one, 1/32.2374

Adding all those up, we notice that we get a geometric series.2378

Our common ratio, what we are multiplying by each time is ½.2383

You got to remember, the sum of a geometric series which we learn back in calculus 2,2388

in level 2 calculus, is the first term in the series ÷ 1 - the common ratio.2393

That is where I got 1/8 is that first term, that is where that came from.2400

The common ratio is that ½.2404

1/8/ 1 - ½ simplifies down to ¼ is my probability.2408

I’m going to calculate this in a different way.2414

Let me show you how you could have done that, in fact without summing up the geometric series.2417

Here is an alternate solution for that one, the same one as before.2423

Alternate solution is to find the probability of A is equal to, we can calculate the complement of A which means the opposite of A.2427

That is the complement A bar means the complement of A, the opposite of what we are looking for.2452

And then do 1 - the probability of the complement of A.2461

Let us figure out what the complement of A is.2466

Remember A was the set of outcomes that have at least 3 flips.2469

That was tail-tail-head, tail-tail-tail-head, and so on.2476

The complement of A is just the outcomes that head 2 or fewer flips.2483

That can be if you get a head right away or you get 1 tail and then a head.2490

That is it, because anything else would have 3 or more flips.2495

The probability of a complement of A is the probability of getting a head immediately +2499

the probability of getting a tail and then a head.2507

The probability of getting a head right away on the first flip is ½.2511

The probability of getting a tail and then a head is ½ × ½.2515

We get ½ + ¼ which is ¾, that was the probability of the complement of A.2524

The probability of A is 1 – that, 1 -3/4 which is ¼.2531

Of course, that is the same answer we got before but using quite a different strategy.2540

It is kind of reassuring.2544

They ought to be the same answer but it is nice to see that they check each other.2546

Let me make sure that is still clear.2553

What we did here was we calculated the compliment of A, that means everything that is not inside A.2557

It is the opposite set from A.2562

A was tail-tail-head, tail-tail-tail-head, all the strings with 3 flips or more.2565

The compliment is all the shorter strings that is just getting a head or getting a tail and a head.2571

The probability of that is, the probably just getting a head on the first flip is ½ because you flip 1 coin.2578

What is the chance of getting a head is ½.2585

The probability of getting a tail and then a head, it would be 1/2 × ½ because you got to get 2 flips,2587

the way you are predicting them.2594

Add those up, you get ¾.2595

We subtract that from 1 because we are using this formula here.2597

1 -3/4 is ¼.2602

That is the probability of the experiment going for at least 3 flips.2605

In our last example here, we are going to draw a 5 card poker hand from a standard 52 card deck.2615

The question is what is the probability that we draw all spades?2621

We want to identify the experiment, all the outcomes in the event that we are interested in.2626

There is a lot of possible outcomes for this experiment.2630

We are going to be learning some new notation in the context of this example2633

because it is going to introduce something that we are going to be studying later.2640

We are going to learn about combinations a little bit in this example.2644

The experiment here is drawing the cards, the set of possible outcomes is all the different ways2648

to draw 5 cards from a standard 52 card deck.2670

This is a little tricky to count.2675

Let me show you how it works out.2677

Think about it, we are drawing from a 52 card deck, there are 52 ways we can draw the first card.2680

After we drawn that first card, there are 51 cards left in the deck.2692

There are 51 choices we can make for the second card.2697

There are 50 choices left for the third card.2703

And then, 49 choices for the 4th and 48 choices for the 5th.2712

That seems like there is this very large number of ways to get a particular a poker hand.2718

However, once you got the 5 cards in your hand, we do not care what order there in.2727

Which means that, you get say the cards you draw are 1, 5, 3, jack, and 8.2733

I’m not even going to worry about the difference suits, the clubs, spades, or hearts, or diamonds.2747

But suppose you get a 1, 5, 3, jack, or 8.2753

You could also had drawn the jack and then the 3 and then the 1 and then the 5 and 8.2758

Once those cards are in your hand, that looks like the same poker hand because you can shuffle them around in your hand.2764

What we did when we are counting all the different ways to draw the 5 cards2772

is we counted these poker hand separately, even though they are really the same poker hand.2776

The problem is they were over counting.2785

We are counting the same poker hand multiple times.2787

We have to take this answer and we have to divide by something and we have to divide by2791

the number of ways that each poker hand got counted.2797

Now, let us just think about a particular poker hand.2803

Let us think about this one in particular.2806

1, 5, 3, jack, 8, let us think about the number of different ways that that poker hand was counted in our original list.2807

Just think about how many different ways could you order those 5 cards.2820

There are 5 choices for which card could have been first, assuming that we are talking about that particular hand.2826

There are 5 ways, the 5 choices for the first card.2834

Once you have picked that first card, there are 4 choices left for the second,2839

and then 3 choices left for the third, and then 2 choices left for the 4th, and then the 5th card would be fixed.2846

Remember, we are talking about a particular poker hand like 1, 5, 3, jack, 8.2854

We are talking those 5 cards, how many different ways could those 5 cards be rearranged2860

and how many different ways that they get counted in this original list.2866

The answer is 5 × 4 × 3 × 2 × 1.2871

We have to divide that out of our list because each poker hand, each group of 5 cards got counted.2878

1 × 2 × 3 × 4 × 5 is 5! ways in our original list.2901

We have to divide out by 5! in order to get an accurate number of the different possible poker hands.2909

That is the total number of outcomes.2919

There is a clever way to simplify this using factorials.2921

Notice that the denominator here is 5!.2926

The numerator is 52 × 51 × 50 × 49 × 48.2933

That looks like we are starting to build up 52!, but what is missing there is 47 × 46, on that down to 1.2942

If I just multiply those on, I have to divide them out again.2953

47 × 46 on down to 1.2957

Another way to write that is as 52! ÷ 47!.2963

That is another way to list the total number of hands.2974

This is actually something we are going to learn about in the next lesson.2978

The total number of ways to choose 5 cards from 52 is, the notation we are going to use is 52 choose 5.2981

There are some other notations along with this notation.2994

It is a little confusing because different textbooks use different notations.2998

C of 52 5, what it means is this factorial formula.3002

This is read as 52 choose 5.3008

What it means is this factorial formula, you expanded it out by doing 52! ÷ 5!, and this 47!3015

really came from the fact that 47 is 52 -5.3026

That is where that came from because that 47 came from this 47 here, which came from this 47,3033

which came from the fact that we had already used up 5 of the factors of 52! there.3039

The number of outcomes here is 52 choose 5.3047

This is our new notation, 52 choose 5.3051

It is the way you calculate it is as 52! ÷ 5! ÷ 47!.3055

Let me recap this because we have to keep going on the next side here.3067

I have not even talked about the event that we are interested in yet.3070

Our experiment is drawing the cards.3073

The number of ways to draw the cards, there are 52 ways to get your first card.3075

Once you draw on that card, there is only 51 card left in the deck, so there are 51 ways to get the second card.3079

There are 50 cards left in the deck, 50 ways, 49 and 48.3086

But the problem is that that over counts because you can get the same poker hand in different possible orders.3090

In order to fix that, we looked in a single poker hand, a typical poker hand,3099

and we figured out that the number of ways that that could have been ordered was 5 × 4 × 3 × 2 × 1.3106

That is why we had to divide out by 5 × 4 × 3 × 2 × 1, it is because each poker hand3113

that counted that many ways in our initial list up here.3120

Each poker hand, we over counted by a factor of 5!.3126

That is our number of outcomes but then as a clever way to write the numerator, we wrote that as 52! ÷ 47!.3132

That is where this part of our answer comes from.3142

The 47 comes from 52 -5 and then this 5! Is where we got that 5! in the denominator there.3145

This is our new notation, 52 choose 5.3157

It is sometimes written with these elongated parentheses.3161

It is sometimes written with this C.3165

C stands for combinations there.3169

By the way, this 52 choose 5, it looks like a fraction but it is not a fraction.3172

It is not the same as 52/5.3176

Do not put the bar there, do not turn it into a fraction because those are totally different things.3180

That is just the number of outcomes, we still need to count all the outcomes in the event that we are interested in.3185

The event was that we draw all spades.3192

Let us try and figure that out.3196

I will try and figure that out in the next slide.3197

But in fact, it would d be fairly easy if you remember the strategy that we used here.3200

Remember that 52 choose 5 is all the ways you can draw 5 cards from 52 cards total.3206

Keeping going with this example, our event is A is ways to get a poker hand, a 5 card poker hand of all spades.3217

Here is the clever way to think about it.3243

Remember how we drew our hand in the first place, the 52 choose 5.3245

Think about if they are all going to be spades.3249

If they are all going to be spades then we are really saying it is the number of ways3254

to choose 5 spade out of how many spades are there total?3263

There are 13, the ace through the king of spades, out of 13.3275

According to our logic before that is exactly 13 choose 5.3283

If we wanted to express that in factorials, that is 13! ÷ 5! ÷ 13 -5! which is 13! ÷ 5! × 8!.3291

That is our number of ways to choose 5 spades out of, there are only 13 in the deck.3310

That is the number of ways that we can get 5 spades.3322

Our probability of A is the number of outcomes in A ÷ the number of outcomes in S.3325

We already calculated both of those, it is 13 choose 5 ÷ the total number of poker hands we can get is 52 choose 5.3351

We could simplify that down if we wanted.3365

13! ÷ 5! × 8!.3367

We already saw above that 52 choose 5 is 52! ÷ 5! × 47!.3373

We could simplify that a bit further.3383

We can cancel the 5! and flip up the denominator.3385

13! ÷ 52!.3389

13! ÷ 8! × 52!.3399

The 47 would flip up to the numerator and the 5! would cancel.3405

I'm going to make no attempt to try to actually calculate that as a number.3413

It would be a very small number.3418

But that is our probability, if you draw 5 cards out of the deck that you will get all 5 cards will be spades.3419

That is of the end of our 6th example there.3434

Let me recap just quickly what we did on the second slide here.3437

Our event was the number ways to get a poker hand of all spades.3441

Using our strategy from before that is 13 choose 5, remember we have learned that notation on the previous slide.3445

If you do not believe that, another way to think about that is to say how many ways can we get all 5 spades.3452

If you are choosing 5 spades, there are 13 ways to get the first one.3464

Once you have picked one spade, there are 12 ways after that to get the second one.3468

11 ways once you picked the first two, × 10 × 9.3473

There is your 5 spades but that over counts because the 5 spades could have come in any order.3478

The number of possible orders they could arrive in is 5 × 4 × 3 × 2 × 1.3485

That simplifies, we have 5! in the denominator and the numerator we can cleverly write it as 13! ÷ 8!.3493

Because we are multiplying all the numbers down from 13 but we are cutting off all the ones down from 8.3506

We can get that as 13! ÷ 8!.3515

That is another way to get to that same answer there for the event that we are interested in.3519

To find the probability, we just take the number of ways to get to our event ÷3525

the total number of outcomes 13 choose 5 ÷ 52 choose 5.3530

We get a kind of cumbersome expression involving factorials in our answer.3535

That wraps up our last example and that wraps up our introductory lecture here on www.educator.com, for probability.3543

We went over some terminology, we practiced some basic probability using counting techniques.3550

On the next set of lectures, we will develop these counting techniques further3556

and we will have complicated probability problems.3560

I hope you will stick around for that.3564

These are the probability lectures here on www.educator.com and my name is Will Murray.3565

Thanks, bye.3570